Problem: Balance the following chemical equation: $ $ $\text{N}_2\text{H}_4 +$ $\text{O}_2 \rightarrow$ $\text{NO}_2 +$ $\text{H}_2\text{O}$
Answer: There are $2 \text{ N}$ on the left and only $1$ on the right, so multiply $\text{NO}_2$ by ${2}$ $ \text{N}_2\text{H}_4 + \text{O}_2 \rightarrow {2}\text{NO}_2 + \text{H}_2\text{O} $ There are $4 \text{ H}$ on the left and only $2$ on the right, so multiply $\text{H}_2\text{O}$ by ${2}$ $ \text{N}_2\text{H}_4 + \text{O}_2 \rightarrow 2\text{NO}_2 + {2}\text{H}_2\text{O} $ That gives us $6 \text{ O}$ on the right and only $2$ on the left, so multiply $\text{O}_2$ by ${3}$ . (Since oxygen is by itself on the left, it should be done at the end because you can give it a coefficient without affecting another element.) $ \text{N}_2\text{H}_4 + {3}\text{O}_2 \rightarrow 2\text{NO}_2 + 2\text{H}_2\text{O} $ The balanced equation is: $ \text{N}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{NO}_2 + 2\text{H}_2\text{O} $